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I am just wondering why a space like $H_0^1(0,\infty)=\{f\in H^1(0,\infty):f(0)=0\}$ is dense in $L^2(0,\infty)$ where $H^1$ is the Sobolev space?

Thanks in advance.

Math

Math
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1 Answers1

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$C^{\infty }$ functions with compact support in $(0,\infty )$ are dense in $L^{2}(0,\infty )$. Such functions belong to $H^{1}_0(0,\infty )$.

Kavi Rama Murthy
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  • A silly quyestion: Why is it that a function in $C^{\infty}$ with compact support must satisfy $f(0)=0$? – Math Jul 20 '18 at 08:07
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    I meant compact support contained in $(0,\infty)$. By the way there is some problem with the statement itself. If your functions are defined on the open interval $(0,\infty )$ what does $f(0)=0$ mean? – Kavi Rama Murthy Jul 20 '18 at 08:14
  • Good point! I saw this in Schmudgen's book on unbounded self-adjoint operators on Page 17. So is this clear to see? I mean if I use $[0,\infty)$, then would it be simple to see that $f(0)=0$? Now, what if I use $H^1(\mathbb R)={f\in L^2(\mathbb R):f'\in L^2(\mathbb R)}$. Would $X={f\in H^1(\mathbb R): f(0)=0}$ still be dense in $L^2(\mathbb R)$? – Math Jul 20 '18 at 08:34
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    It would be. You can approximate $f\in X$ by smooth functions with compact support contained in $\mathbb R \setminus {0}$. (These function actually vanish in a neighborhood of $0$). – Kavi Rama Murthy Jul 20 '18 at 08:40
  • Great! I will try to do that and see what I can get... – Math Jul 20 '18 at 08:42